Calculate the volume 3.00 moles of a gas will occupy at 24.0 °C and 101.3 kPa. 2. How many moles of gas would be present in a gas trapped within a 100.0 mL vessel at 25.0 °C at a pressure of 2.50 atmospheres? 3. If the number of moles of a gas are doubled at the same temperature and pressure, will the volume increase or decrease?
What size metal duct should be used to deliver 270 CFM with a pressure drop of 0.15 in wc if the total equivalent length is 80 ft? (57.10) 8 in 5. What is the velocity of 500 CFM of air moving through a 10 in duct? (57.10) 900 fpm. 6.
(1) 2. The temperature of a 2.0 g sample of aluminium increases from 25°C to 30°C. How many joules of heat energy were added? (Specific heat of Al = 0.90 J g−1K−1) A. 0.36 B.
(b) Calculate the volume of 0.2M UO3- needed to react with 20.00 cm3 of 0.1M Cr2O72-. 3. 24.40 g of hydrated iron(II) sulphate, FeSO4.xH2O was dissolved and made up to 1.0 dm3 of aqueous solution, acidified with sulphuric acid. 25.00 cm3 of the solution was titrated with 20.00 cm3 of 0.022M potassium manganate(VII) solution for complete oxidation. a) Write the equation for the reaction.
Finish by recording volume and temperature. III. Observations and Data Atmospheric Pressure | 769 mmHg | Height of water above glass | 50 mmH2O | Number of moles of air in the graduated cylinder: 5.957 x 10-4 Temperature | Volume | Adjusted Volume | Air’s Partial Pressure | Water Vapor Pressure | 80°C | 125 mL | 124.8 mL | 105.2 mmHg | 720.3 mmHg | 75°C | 115 mL | 114.8 mL | 112.72 mmHg | 712.3 mmHg | 70°C | 100 mL | 99.8 mL | 127.8 mmHg | 697.7 mmHg | 65°C | 97 mL | 96.8 mL | 129.9 mmHg | 695.6 mmHg | 60°C | 87.5 mL | 87.3 mL | 141.9 mmHg | 683.6 mmHg | 55°C | 79 mL | 78.8mL | 163.1 mmHg | 662.4 mmHg | 50°C | 75 mL | 74.8 mL | 160.6 mmHg | 664.9 mmHg | 0°C | 12.5 mL | 12.3 mL | 825.5 mmHg | 0 mmHg | Pressure of Cylinder | Pressure of Atmosphere | Pressure Exerted by water | 825.5 mmHg | 769 mmHg | 56.54 mmHg | * Percent error: * Hexane is completely non-polar and will readily interact with other molecules, while water is polar and will not interact as easily. Therefore, hexane is more likely to break its bond to
NaCl 1 Na+ and 1 Cl- are formed). Calculate the osmolarity of that solution. What is the molar concentration of ALL particles? e.g. 0.15 M NaCl solution = 0.15 moles of Na+ atoms + 0.15 moles of Cl- atoms = 0.30 Osmoles In other words, the solution is said to have an osmolarity of 0.30 Osm (or 300 mOsm) Assume the osmolarity of the ICF of body cells to be 0.300 Osm (300 mOsm) 2nd, determine if the solute is a PENETRATING particle or is NON-PENETRATING.
The system was heated for 4-5 hr under vacuum at 200°C and then cooled down to the temperature (50°C) where we want to perform the adsorption study. Small doses of test gases consecutively introduced to the system and gradually increased up to 50 Torr until an equilibrium pressure was reached. Then the obtained differential heats of the test gases adsorption were recorded as a function of its coverage. Further, the manifold degassed under vacuum for almost 30 minutes; adsorption was conducted in the same manner. Finally, the number and strength of active surface are obtained from the difference between the adsorbed gases from the first and second
of boiling water | 100°C | 100°C | Peak temp. the water reaches | 22°C | 24°C | How much change in degrees | Δ2°C | Δ3°C | Pat and Jaymie’s Data | Trial #1 | Trial #2 | Mass of zinc | 6.0g | 7.99g | Mass of water in
40 x 0.25 =10.0. If the volume was 2.5 mL the pressure would be 4.0 atm. 2.5 x 4.0 = 10.0. The number of moles of the gas and the temperature are assumed to be constant in this experiment. PV=k.
When the volume is halved from 20.0 mL to 10.0 mL the pressure is almost doubled from 53.22 kPa to 104.54 kPa. The relationship between the pressure and volume of a confined gas is inverse. The line of the graph is curved not straight. This indicates an indirect relationship. The amount of the gas and the temperature are assumed to be constant in this experiment.